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# Tang 02 enthalpy and hess' law

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### Tang 02 enthalpy and hess' law

1. 1. ENTHALPY AND HESS’ LAW
2. 2. Chemical energy is stored in: – moving electrons – vibration of chemical bonds – rotation and translation of molecules – stored nuclear energy of protons & neutrons – energy stored in chemical bonds ENTHALPY AND HESS’ LAW
3. 3. Enthalpy enthalpy (H) – total kinetic and potential energy of a system at a constant pressure change in enthalpy (∆H) – change in heat of a system ∆H = Hfinal - Hinitial ∆H = Hproducts - Hreactants ENTHALPY AND HESS’ LAW
4. 4. Enthalpy For an exothermic reaction: Hproducts < Hreactants ∴∆H is negative Describe what has happened. H reaction progress reactants products ENTHALPY AND HESS’ LAW
5. 5. Enthalpy For an endothermic reaction: Hproducts > Hreactants ∴∆H is positive Describe what has happened. H reaction progress reactants products ENTHALPY AND HESS’ LAW
6. 6. Enthalpy How does enthalpy relate to heat with respect to an isolated system? ∆H = q when using a calorimeter. What is the formula for q? ENTHALPY AND HESS’ LAW
7. 7. Enthalpy What is the formula for q? q = mcΔT Since q = ΔH ΔH = mcΔT ENTHALPY AND HESS’ LAW
8. 8. Enthalpy How are enthalpy and heat related? ∆Hsystem = -q exothermic reaction ∆Hsystem = +q endothermic reaction This is only true at constant pressure, which is always the case in a calorimeter. Standard conditions for enthalpy changes: temperature at 298K ENTHALPY AND HESS’ LAW
9. 9. Enthalpy The enthalpy change for a balanced chemical reaction is constant. e.g. 2 H2(g) + O2(g)  2 H2O(g) ∆H° = - 483.6 kJ ∆H° – standard enthalpy of a reaction – must be a balanced chemical equation Read as - 483.6 kJ per 2 mol of H2 being reacted. ENTHALPY AND HESS’ LAW This is called a thermochemical equation
10. 10. Enthalpy ∆H° values are proportional to coefficient changes of a chemical equation. N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ 3 N2O4 (g)  6 NO2 (g) ∆H° = +173.79 kJ ½ N2O4 (g)  NO2 (g) ∆H° = +28.96 kJ ENTHALPY AND HESS’ LAW
11. 11. Enthalpy N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ 2 NO2 (g)  N2O4(g) ∆H° = -57.93 kJ Reversing a chemical reaction causes a sign change in front of the ∆H° value. ENTHALPY AND HESS’ LAW
12. 12. Enthalpy ∆H° values of unknown reactions can be solved when other known reactions are given. ENTHALPY AND HESS’ LAW
13. 13. Example #1 C(s) + O2(g)  CO2(g) ∆H° = ? C(s) + ½ O2(g)  CO(g) ∆H° = -110.5 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW What parts of the above reactions are the same as the first reaction?
14. 14. Rules for solving thermochemical equation questions: For any reaction that can be written in steps, the ∆H° is the same as the sum of the values of the ∆H° for each individual step. Hess’s Law of Summation ENTHALPY AND HESS’ LAW
15. 15. ∆H° Rules 1. If all the coefficients of an eqn are multiplied or divided by a common factor, the ∆H° must be changed likewise. ENTHALPY AND HESS’ LAW N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ 3 N2O4 (g)  6 NO2 (g) ∆H° = +57.93kJ x 3 = +173.79 kJ ½ N2O4 (g)  NO2 (g) ∆H° = +57.93kJ ÷ 2 = +28.96 kJ
16. 16. ∆H° Rules 2. When a reaction is reversed, the sign of ∆H° must also be reversed. ENTHALPY AND HESS’ LAW N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ 2 NO2 (g)  N2O4(g) ∆H° = -57.93 kJ
17. 17. ∆H° Rules 3. When canceling compounds for Hess’s Law, the state of the compounds is important. ENTHALPY AND HESS’ LAW H2(g) + ½ O2(g)  H2O(g) ΔH° = +57.93 kJ K2SO4(s) + 2H2O(l)  H2SO4(aq) + 2KOH(s) ΔH° = +342.4kJ These two CANNOT cancel each other out
18. 18. C(s) + O2(g)  CO2(g) ∆H° = ? C(s) + ½ O2(g)  CO(g) ∆H° = -110.5 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW Ex #1 C(s) + O2(g)  CO2(g) ∆H° = -393.5 kJ The enthalpies are added together Therefore the enthalpy is -393.5 kJ
19. 19. 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ? Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW Ex #2 This is not as easy to do! STEP 1: Number each given thermochemical equation. 1 2
20. 20. 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ? Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW Ex #2 1 2 STEP 2: Arrange the equations so that your desired reactants are on the left side, and your desired products are on the right side 2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g) ∆H° = +26.7 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ 1- 2
21. 21. 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ? Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW Ex #2 1 2 2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g) ∆H° = +26.7 kJ 3CO(g) + 3/2 O2(g)  3CO2(g) ∆H° = -849.0 kJ 1- 2 STEP 3: Multiply the equations by factors such that they may match your desired equation. Remember to multiply the enthalpy by the same factor. 3 x
22. 22. 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ? Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ ENTHALPY AND HESS’ LAW Ex #2 1 2 2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g) ∆H° = +26.7 kJ 3CO(g) + 3/2 O2(g)  3CO2(g) ∆H° = -849.0 kJ 1- 2 STEP 4: Add the two equations and enthalpies together. Remember to cancel repeating chemicals. 3 x 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = -822.3 kJ
23. 23. Enthalpy H2O (s)  H2O (l) What is happening? Is a ∆H° value involved? Why or why not? ENTHALPY AND HESS’ LAW Heat is absorbed by the ice in order to melt. ΔH° is positive, because heat is required
24. 24. Enthalpy Enthalpy is also involved in physical changes. ENTHALPY AND HESS’ LAW